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Help me figure out this logic problem, it seems impossible~!

Ok a basic run down of the problem.


You've got 5 people, they have to go across this bridge in 30 seconds. They need to go in ones or twos. They have to carry a lamp with them at all times when crossing the bridge. They must cross at the slowest one in the groups speed. They cross at these speeds

1 second
3 seconds
6 seconds
8 seconds
12 seconds


So logic would make you think have the 1 second guy go across with everyone and come back as fast as he could. But that doesn't work.
 
Originally posted by: Cuda1447
Ok a basic run down of the problem.


You've got 5 people, they have to go across this bridge in 30 seconds. They need to go in ones or twos. They have to carry a lamp with them at all times when crossing the bridge. They must cross at the slowest one in the groups speed. They cross at these speeds

1 second
3 seconds
6 seconds
8 seconds
12 seconds


So logic would make you think have the 1 second guy go across with everyone and come back as fast as he could. But that doesn't work.
Click to expand...


yea thats what I tried too....
 
Hint: The trick to any bridge crossing problems like this is that you need to get the slowest people to go across together, effectively removing the second slowest person from the equation. I'll let you try it with that hint.
 
Originally posted by: Basilisk6
Hint: The trick to any bridge crossing problems like this is that you need to get the slowest people to go across together, effectively removing the second slowest person from the equation. I'll let you try it with that hint.
Click to expand...

still doesn't work! Just gimme the solution if your so smart lol
 
I've seen many of these type of problems. The trick is to take the two fastest guys across, one comes back with lamp. then the two slowest guys go accross, and the other fastest guy bring the lamp back. that move saves alot of time.
 
Originally posted by: Basilisk6
Hint: The trick to any bridge crossing problems like this is that you need to get the slowest people to go across together, effectively removing the second slowest person from the equation. I'll let you try it with that hint.
Click to expand...

doesnt work
 
Spoiler:

























First, send the 1 and 3 across. Send the 1 back. Send the 1 and the 6 across. Send the 1 back. Send the 12 and 8 across, and send the 3 back. Finally send the 3 and 1 across. Done in 29 seconds.

3 + 1 + 6 + 1 + 12 + 3 + 3 = 29
 
Originally posted by: Gobadgrs
Originally posted by: Basilisk6
Hint: The trick to any bridge crossing problems like this is that you need to get the slowest people to go across together, effectively removing the second slowest person from the equation. I'll let you try it with that hint.
Click to expand...

doesnt work
Click to expand...

Sure it does. Solution:














Send 1 and 3 (3 seconds)
Send 1 (4 seconds)
*Send 8 and 12 (16 seconds)*
Send 3 (19 seconds)
Send 1 and 6 (25 seconds)
Send 1 (26 seconds)
Send 1 and 3 (29 seconds)

And you're done with 1 second to spare.
 
Originally posted by: lozina
I've seen many of these type of problems. The trick is to take the two fastest guys across, one comes back with lamp. then the two slowest guys go accross, and the other fastest guy bring the lamp back. that move saves alot of time.
Click to expand...

Ahh ok, I figured it out now. lol
 
****SPOLIER*****


























1) 1 and 3, send 1 back
2) 8 and 12, send 3 back
3)6 and 1, send 1 baclk
4)1 and 3.

Finish with 29 seconds.
 
got it!

send over 1 and 3, then send back 1, send over 8 and 12 then send back 3 send over 1 and 5 then send back 1 then send 1 and 3
 
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